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ML Aggarwal Solutions for Class 8 Maths

Chapter 10: Algebraic Expressions and Identities

Exercise 10.1 1. Identify the terms, their numerical as well as literal coefficients in each of the following

expressions:

(i) 12x2yz – 4xy2

(ii) 8 + mn + nl – lm

(iii) x2/3 + y/6 – xy2

(iv) -4p + 2.3q + 1.7r

Solution:

2. Identify monomials, binomials, and trinomials from the following algebraic expressions :

(i) 5p × q × r2

(ii) 3x2 + y ÷ 2z

(iii) -3 + 7x2

(iv) (5a2 – 3b2 + c)/2

(v) 7x5 – 3x/y

(vi) 5p ÷ 3q – 3p2 × q2

Solution:

(i) 5p × q × r2 = 5pqr2

As this algebraic expression has only one term, its therefore a monomial.

(ii) 3x2 + y ÷ 2z = 3x2/2z + y/2z

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As this algebraic expression has two terms, its therefore a binomial.

(iii) -3 + 7x2


(iv)

As this algebraic expression has three terms, its therefore a trinomial.

(v) 7x5 – 3x/y


(vi) 5p ÷ 3q – 3p2 × q2 = 5p/3q - 3p2q2


3. Identify which of the following expressions are polynomials. If so, write their degrees.

(i) 2/5x4 – √3x2 + 5x – 1

(ii) 7x3 – 3/x2 + √5

(iii) 4a3b2 – 3ab4 + 5ab + 2/3

(iv) 2x2y – 3/xy + 5y3 + √3

Solution:

(i) It is a polynomial and the degree of this expression is 4.

(ii) It is not a polynomial.

(iii) It is a polynomial and the degree of this expression is 5.

(iv) It is not a polynomial.

4. Add the following expressions:

(i) ab – bv, bv – ca, ca – ab

(ii) 5p2q2 + 4pq + 7, 3 + 9pq – 2p2q

(iii) l2 + m2 + n2, lm + mn, mn + nl, nl + lm

(iv) 4x3 – 7x2 + 9, 3x2 – 5x + 4, 7x3 – 11x + 1, 6x2 – 13x

Solution:

(i) ab – bc, bc – ca, ca – ab

On adding the expressions, we have

⇒ ab – bc + bc – ca + ca – ab = 0

(ii) 5p2q2 + 4pq + 7,3 + 9pq – 2p2q2


= 5p2q2 + 4pq + 7 + 3 + 9pq – 2p2q2

= 5p2q2 – 2p2q2 + 4pq + 9pq + 7 + 3

= 3p2q2 + 13pq + 10

(iii) l2 + m2 + n2, lm + mn, mn + nl, nl + lm






= l2 + m2 + n2 + lm + mn + mn + nl + nl + lm

= l2 + m2 + n2 + 2lm + 2mn + 2nl

(iv) 4x3 – 7x2 + 9, 3x2 – 5x + 4, 7x3 – 11x + 1, 6x2 – 13x


= 4x3 – 7x2 + 9 + 3x2 – 5x + 4 + 7x3 – 112 + 1 + 6x2 – 13x

= 4x2 + 7x3 – 7x2 + 3x2 + 6x2 – 5x – 11x – 13x + 9 + 4 + 1

= 11x3 – 2x2 – 29x + 14

5. Subtract:

(i) 8a + 3ab – 2b + 7 from 14a – 5ab + 7b – 5

(ii) 8xy + 4yz + 5zx from 12xy – 3yz – 4zx + 5xyz

(iii) 4p2q – 3pq + 5pq2 – 8p + 7q -10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q

Solution:

(i) Subtracting 8a + 3ab – 2b + 7 from 14a – 5ab + 7b – 5, we have

= (14a – 5ab + 7b – 5) – (8a + 3ab – 2b + 7)

= 14a – 5ab + 7b – 5 – 8a – 3ab + 2b – 7

= 6a – 8ab + 9ab – 12

(ii) Subtracting 8xy + 4yz + 5zx from 12xy – 3yz – 4zx + 5xyz, we have

= (12xy – 3yz – 4zx + 5xyz) – (8xy + 4yz + 5zx)

= 12xy – 3yz – 4zx + 5xyz – 8xy – 4yz – 5zx

= 4xy – 7yz – 9zx + 5xyz

(iii) Subtracting 4p2q – 3pq + 5pq2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq2 + 5p2q, we have

= (18 – 3p – 11q + 5pq – 2pq2 + 5p2q) – (4p2q – 3pq + 5pq2 – 8p + 7q – 10)

= 18 – 3p – 11q + 5pq – 2pq2 + 5p2q – 7p2q + 3pq – 5pq2 + 8p – 7q + 10

= 28 + 5p – 78q + 8pq – 7pq2 + p2q

6. Subtract the sum of 3x2 + 5xy + 7y2 + 3 and 2x2 – 4xy – 3y2 + 7 from 9x2 – 8xy + 11y2

Solution:

First, adding 3x2 + 5xy + 7y2 + 3 and 2x2 – 4xy – 3y2 + 7, we have

= 3x2 + 5xy + 7y2 + 3 + 2x2 – 4xy – 3y2 + 7

= 5x2 + xy + 4y2 + 10

Now,

Subtracting 5x2 + xy + 4y2 + 10 from 9x2 – 8xy + 11y2

= (9x2 – 8xy + 11y2) – (5x2 + xy + 4y2 + 10)

= 9x2 – 8xy + 11y2 – 5x2 – xy – 4y2 – 10

= 4x2 – 9xy + 7y2 – 10

7. What must be subtracted from 3a2 – 5ab – 2b2 – 3 to get 5a2 – 7ab – 3b2 + 3a?

Solution:

From the question, its understood that we have to subtract 5a2 – 7ab – 3b2 + 3a from 3a2 – 5ab – 2b2 – 3





= 3a2 – 5ab – 2b2 – 3 – (5a2 – 7ab – 3b2 + 3a)

= 3a2 – 5ab – 2b2 – 3 – 5a2 + 7ab + 3b2 – 3a

= -2a2 + 2ab + b2 – 3a – 3

8. The perimeter of a triangle is 7p2 – 5p + 11 and two of its sides are p2 + 2p – 1 and 3p2 – 6p + 3.

Find the third side of the triangle.

Solution:

Given,

Perimeter of a triangle = 7p2 – 5p + 11

And, two of its sides are p2 + 2p – 1 and 3p2 – 6p + 3

We know that,

Perimeter of a triangle = Sum of three sides of triangle

⇒ 7p2 – 5p + 11 = (p2 + 2p – 1) + (3p2 – 6p + 3) + (Third side of triangle)

7p2 – 5p + 11 = (4p2 – 4p + 2) + (Third side of triangle)

⇒ Third side of triangle = (7p2 – 5p + 11) – (4p2 – 4p + 2)

= (7p2 – 4p2) + (- 5p + 4p) + (11 – 2)

= 3p2 – p + 9

Thus, the third side of the triangle is 3p2 – p + 9.





Exercise 10.2

1. Find the product of:

(i) 4x3 and -3xy

(ii) 2xyz and 0

(iii) –(2/3)p2q, (3/4)pq2 and 5pqr

(iv) -7ab, -3a3 and –(2/7)ab2

(v) –½x2 – (3/5)xy, (2/3)yz and (5/7)xyz

Solution:

Product of:

(i) 4x3 and -3xy = 4x3 × (-3xy) = -12x3+1 y = -12x4y

(ii) 2xyz and 0 = 2xyz × 0 = 0

2. Multiply:

(i) (3x – 5y + 7z) by – 3xyz

(ii) (2p2 – 3pq + 5q2 + 5) by – 2pq

(iii) (2/3a2b – 4/5ab2 + 2/7ab + 3) by 35ab

(iv) (4x2 – 10xy + 7y2 – 8x + 4y + 3) by 3xy

Solution:

(i) – 3xyz × (3x – 5y + 7z)

= (- 3xyz) × 3x + (- 3xyz) × (- 5y) + (- 3xyz) × (7z)

= – 9x2yz + 15xyz2 – 21xyz2

(ii) -2pq × (2p2 – 3pq + 5q2 + 5)





= (-2pq) × 2p2 + (-2pq) × (-3pq) + (- 2pq) × (5q2) + (-2pq) × 5

= -4p3q + 6p2q2 – 10pq3 – 10pq

(iii) by 35ab

= (2/3)a2b × 35ab – (4/5)ab2 × 35ab + (2/7)ab × 35ab + 3 × 35ab

= (70/3)a3b2 – 28a2b3 + 10a2b2 + 105ab

(iv) (4x2 – 10xy + 7y2 – 8x + 4y + 3) by 3xy

= 4x2 × 3xy – 10xy × 3xy + 7y2 × 3xy – 8x × 3xy + 4y × 3xy + 3 × 3xy

= 12x3y – 30x2y2 + 21xy3 – 24x2y + 12xy2 + 9xy

3. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths

respectively:

(i) (p2q, pq2)

(ii) (5xy, 7xy2)

Solution:

(i) Given, sides of a rectangle are p2q and pq2

Hence,

Area = p2q × pq2 = p2+1 × q2+1 = p3q3

(ii) Given, sides are 5xy and 7xy2

Hence,

Area = 5xy × 7xy2 = 35x1+1 × y1+2 = 35x2y3

4. Find the volume of rectangular boxes with the following length, breadth and height

respectively:

(i) 5ab, 3a2b, 7a4b2

(ii) 2pq, 4q2, 8rp

Solution:

Given are the length, breadth and height of a rectangular box:

(i) 5ab, 3a2b, 7a4b2

∴ Volume = Length × breadth × height

= 5ab × 3a2b × 7a4b2

= 5 × 3 × 7 × a1+2+4 × b1+1+2

= 105a7b4

(ii) 2pq, 4q2, 8rp

∴ Volume = Length × breadth × height

= 2pq × 4q2 × 8rp

= 2 × 4 × 8 × p1+1 × q1+2 × r

= 64p2q3r

5. Simplify the following expressions and evaluate them as directed:





(i) x2(3 – 2x + x2) for x = 1; x = -1; x = 2/3 and x = –1/2

(ii) 5xy(3x + 4y – 7) – 3y(xy – x2 + 9) – 8 for x = 2, y = -1

Solution:

(i) x2(3 – 2x + x2)

For x = 1; x = -1; x = 2/3 and x = –1/2

x2(3 – 2x + x2) = 3x2 – 2x3 + x4

(a) For x = 1

3x2 – 2x3 + x4 = 3(1)2 – 2(1 )3 + (1)4

= 3 × 1 – 2 × 1 + l

= 3 – 2 + 1 = 2

(b) For x = -1

3x2 – 2x3 + x4 = 3(-1)2 – 2(-1)3 + (-1)4

= 3 × 1 – 2 × (-1) + 1

= 3 + 2 + 1 = 6

(c) For x = 2/3

3x2 – 2x3 + x4 = 3(2/3)2 – 2(2/3)3 + (2/3)4

= 3 × (4/9) – 2 × (8/27) + (16/81)

= (4/3) – (16/27) + (16/81)

= (108 – 48 + 16)/81

= (124 - 48)/81

= 76/81

(d) For x = -1/2

3x2 – 2x3 + x4 = 3(-1/2)2 – 2(-1/2)3 + (-1/2)4

= 3 × (1/4) – 2 × (-1/8) + (1/16)

= (3/4) + ¼ + (1/16)

= (12 + 4 + 1)/16

= 17/16

(ii) 5xy(3x + 4y – 7) – 3y(xy – x2 + 9) – 8

= 15x2y + 20xy2 – 35xy – 3xy2 + 3 x2y – 21y – 8

= 18x2y + 17xy2 – 35xy – 27y – 8

When x = 2, y = -1, we have

= 18(2)2 × (-1) + 17(2) (-1)2 – 35(2) (-1) – 27(-1) – 8

= 18 × 4 × (-1) + 17 × 2 × 1 – 35 × 2 × (-1) – 27 × (-1) – 8

= -74 + 34 + 70 + 27 – 8

= 131 – 80 = 51

6. Add the following:

(i) 4p(2 – p2) and 8p3 – 3p

(ii) 7xy(8x + 2y – 3) and 4xy2(3y – 7x + 8)

Solution:

Adding,

(i) 4p(2 – p2) and 8p3 – 3p





= 8p – 4p3 + 8p3 – 3p

= 5p + 4p3

= 4p3 + 5p

(ii) 7xy(8x + 2y – 3) and 4xy2(3y – 7x + 8)

= 56x2y + 14xy2 – 21xy + 12xy3 – 28x2y2 + 32xy2

= 12xy3 – 28x2y2 + 56x2y +46xy2 – 21xy

7. Subtract:

(i) 6x(x – y + z)- 3y(x + y – z) from 2z(-x + y + z)

(ii) 7xy(x2 -2xy + 3y2) – 8x(x2y – 4xy + 7xy2) from 3y(4x2y – 5xy + 8xy2)

Solution:

Subtracting,

(i) 6x(x – y + z) – 3y(x + y – z) from 2z(-x + y + z)

⇒ 6x2 – 6xy + 6xz – 3xy – 3y2 + 3yz from – 2xz + 2yz + 2z2

= (-2xz + 2yz + 2z2) – (6x2 – 6xy + 6xz – 3xy – 3y2 + 3yz)

= – 2xz + 2yz + 2z2 – 6x2 + 6xy – 6xz + 3xy + 3y2 – 3yz

= 9xy – yz – 8zx – 6x2 + 3y2 + 2z2

(ii) 7xy(x2 – 2xy + 3y2) – 8x(x2y – 4xy + 7xy2) from 3y(4x2y – 5xy + 8xy2)

⇒ 7x3y – 14x2y2 + 21xy3 – 8x3y + 32x2y – 56x2y2 from 12x2y2 – 15xy2 + 24xy3

= (12x2y2 – 15xy2 + 24xy3) – (7x3y – 14x2y2 + 21xy3 – 8x3y + 32x2y – 56x2y2

= 12x2y2 – 15xy2 + 24xy3 – 7x3y + 14x2y2 – 12xy3 + 8x3y – 32x2y + 56x2y2

= 82x2y2 + 3xy3 + x3y – 15xy2 – 32x2y





Exercise 10.3

1. Multiply:

(i) (5x – 2) by (3x + 4)

(ii) (ax + b) by (cx + d)

(iii) (4p – 7) by (2 – 3p)

(iv) (2x2 + 3) by (3x – 5)

(v) (1.5a – 2.5b) by (1.5a + 2.56)

(vi)

Solution:

(i) (5x – 2) by (3x + 4)

= (5x – 2) × (3x + 4)

= 5x (3x + 4) – 2 (3x + 4)

= 15x2 + 20x – 6x – 8

= 15x2+ 14x – 8

(ii) (ax + b) by (cx + d)

= (ax + b) × (cx + d)

= ax (cx + d) + b (cx + d)

= acx2 + adx + bcx + bd

(iii) (4p – 7) by (2 – 3p)

= (4p – 7) × (2 – 3p)

= 4p(2 – 3p) -7(2 – 3p)

= 8p – 12p2 – 14 + 21p

= 29p – 12p2 – 14

(iv) (2x2 + 3) by (3x – 5)

= (2x2 + 3) (3x – 5)

= 2x2(3x – 5) + 3(3x – 5)

= 6x3 – 10x2 + 9x – 15

(v) (1.5a – 2.5b) by (1.5a + 2.5b)

= (1.5a – 2.5b) (1.5a + 2.5b)

= 1.5a(1.5 + 2.5b) – 2.5b(1.5a + 2.5b)

= 2.25a2 + 3.75ab – 3.75a6 – 6.25b2

= 2.25a2 – 6.25b2





2. Multiply:

(i) (x – 2y + 3) by (x + 2y)

(ii) (3 – 5x + 2x2) by (4x – 5)

Solution:

(i) (x – 2y + 3) by (x + 2y)

= (x – 2y + 3) × (x + 2y)

= x (x + 2y) – 2y(x + 2y) + 3 (x + 2y)

= x2 + 2xy – 2xy – 4y2 + 3x + 6y

= x2 – 4y2 + 3x + 6y

(ii) (3 – 5x + 2x2) by (4x – 5)

= (4x – 5) (3 – 5x + 2x2)

= 4x(3 – 5x + 2x2) – 5(3 – 5x + 2x2)

= 12x – 20x2 + 8x3 – 15 + 25x – 10x2

= 8x3 – 30x2 + 37x – 15

3. Multiply:

(i) (3x2 – 2x – 1) by (2x2 + x – 5)

(ii) (2 – 3y – 5y2) by (2y – 1 + 3y2)

Solution:

(i) (3x2 – 2x – 1) by (2x2 + x – 5)

= (3x2 – 2x – 1) (2x2 + x – 5)





= 3x2(2x2 + x – 5) – 2x(2x2 + x – 5) -1(2x2 + x – 5)

= 6x4 + 3x3 – 15x2 – 4x3 – 2x2 + 10x – 2x2 – x + 5

= 6x4 – x3 – 19x2 + 9x + 5

(ii) (2 – 3y – 5y2) by (2y- 1 + 3y2)

= (2 – 3y – 5y2) × (2y- 1 + 3y2)

= 2(2y – 1 + 3y2 ) - 3y (2y – 1 + 3y2) -5y2(2y – 1 + 3y2)

= 4y – 2 + 6y2 – 6y2 + 3y – 9y3 – 10y3 + 5y2 – 15y4

= -15y4 – 19y3 + 5y2 + 7y – 2

4. Simplify:

(i) (x2 + 3) (x – 3) + 9

(ii) (x + 3) (x – 3) (x + 4) (x – 4)

(iii) (x + 5) (x + 6) (x + 7)

(iv) (p + q – 2r) (2p – q + r) – 4qr

(v) (p + q) (r + s) + (p – q)(r – s) – 2(pr + qs)

(vi) (x + y + z) (x – y + z) + (x + y – z) (-x + y + z) – 4zx

Solution:

(i) (x2 + 3) (x – 3) + 9

= x2 (x – 3) + 3(x – 3) + 9

= x2 – 3x2 + 3x – 9 + 9

= x3 – 3x2 + 3x

(ii) (x + 3) (x – 3) (x + 4) (x – 4)

= {(x + 3) (x – 3)} × {(x + 4) (x – 4)}

= {x (x – 3) + 3 (x – 3)} {x (x – 4) + 4 (x – 4)}

= (x2 – 3x + 3x – 9) {x2 – 4x + 4x – 16}

= (x2 – 9) (x2 – 16)

= x2 (x2 – 16) – 9 (x2 – 16)

= x4 – 16x2 – 9x2 + 144

= x4 – 25x2 + 144

(iii) (x + 5) (x + 6) (x + 7)

= {(x + 5) × (x + 6)} (x + 7)

= (x2 + 6x + 5x + 30) (x + 7)

= (x2 + 11x + 30) (x + 7)

= x(x2+ 11x + 30) + 7(x2+ 11x + 30)

= x3 + 11x2 + 30x + 7x2 + 77x + 210

= x3 + 18x2 + 107x + 210

(iv) (p + q – 2r)(2p – q + r) – 4qr

= p(2p – q + r) + q(2p – q + r) – 2r(2p – q + r) – 4qr

= 2p2 – pq + pr + 2pq – q2 + qr – 4pr + 2qr – 2r2 – 4qr

= 2p2 – q2 – 2r2 + pq – 3pr – 2qr





(v) (p + q)(r + s) + (p – q) (r – s) – 2(pr + qs)

= (pr + ps + qr + qs) + (pr – ps – qr + qs) – 2pr – 2qs

= 0

(vi) (x + y + z)(x – y + z) + (x + y – z)(-x + y + z) – 4zx

= x2 – xy + xz + xy – y2 + yz + xz – yz + z2 – x2 + xy + xz

– xy + x2 + yx + xz – yz – z2 – 4zx

= 0

5. If two adjacent sides of a rectangle are 5x2 + 25xy + 4y2 and 2x2 – 2xy + 3y2, find its area.

Solution:

Given,

The adjacent sides of a rectangle are 5x2 + 25xy + 4y2 and 2x2 – 2xy + 3y2

So,

Area of rectangle = Product of two adjacent sides

= (5x2 + 25xy + 4y2) (2x2 – 2xy + 3y2)

= 10x4– 10x3y+ 15x2y2 + 50x3y – 50x2y2 + 75xy3 + 8x2y2 – 8xy3 + 12y4

= 10x4 + 40x3y – 27x2y2 + 67xy3 + 12y4

Thus,

The area of the rectangle is 10x4 + 40x3y – 27x2y2 + 67xy3 + 12y4.





Exercise 10.4

1. Divide:

(i) – 39pq2r5 by – 24p3q3r

(ii) –a2b3 by a3b2

Solution:

(i) – 39pq2r5 (÷) – 24p3q3r

= – 39pq2r5/ – 24p3q3r

2. Divide:

(i) 9x4 – 8x3 – 12x + 3 by 3x

(ii) 14p2q3 – 32p3q2 + 15pq2 – 22p + 18q by – 2p2q.

Solution:





3. Divide:

(i) 6x2 + 13x + 5 by 2x + 1

(ii) 1 + y3 by 1 + y

(iii) 5 + x – 2x2 by x + 1

(iv) x3 – 6x2 + 12x – 8 by x – 2

Solution:

(i) 6x2 + 13x + 5 ÷ 2x + 1

∴ Quotient = 3x + 5 and remainder = 0

(ii) 1 + y3 ÷ 1 + y





∴ Quotient = y2 – y + 1 and remainder = 0

(iii) On arranging the terms of dividend in descending order of powers of x and then dividing, we get

– 2x2 + x + 5 ÷ x + 1

∴ Quotient = – 2x + 3 and remainder = 2

(iv) x3 – 6x2 + 12x – 8 ÷ x – 2

∴ Quotient = x2 – 4x + 4 and remainder = 0

4. Divide:

(i) 6x3 + x2 – 26x – 25 by 3x – 7

(ii) m3 – 6m2 + 7 by m – 1

Solution:

(i) 6x3 + x2 – 26x – 25 ÷ 3x – 7





∴ Quotient = 2x2 + 5x + 3 and remainder = – 4

(ii) m3 – 6m2 + 7 ÷ m – 1

∴ Quotient = m2 – 5m – 5 and remainder = 2.

5. Divide:

(i) a3 + 2a2 + 2a + 1 by a2 + a + 1

(ii) 12x3 – 17x2 + 26x – 18 by 3x2 – 2x + 5

Solution:

(i) a3 + 2a2 + 2a + 1 ÷ a2 + a + 1





∴ Quotient = a + 1 and remainder = 0.

(ii) 12x3 – 17x2 + 26x – 18 ÷ 3x2 – 2x + 5

∴ Quotient = 4x – 3 and remainder = -3

6. If the area of a rectangle is 8x2 – 45y2 + 18xy and one of its sides is 4x + 15y, find the length of

adjacent side.

Solution:

Given,

Area of rectangle = 8x2 – 45y2 + 18xy

And, one side = 4x + 15y

∴ Second (adjacent) side = Area of rectangle/ One side

= 8x2 – 45y2 + 18xy ÷ 4x + 15y

Thus, length of the adjacent side is 2x – 3y.





Exercise 10.5

1. Using suitable identities, find the following products:

(i) (3x + 5) (3x + 5)

(ii) (9y – 5) (9y – 5)

(iii) (4x + 11y) (4x – 11y)

(iv) (3m/2 + 2n/3) (3m/2 – 2n/3)

(v) (2/a + 5/b) (2a + 5/b)

(vi) (p2/2 + 2/q2) (p2/2 – 2/q2)

Solution:

(i) (3x + 5) (3x + 5)

= (3x + 5)2

= (3x)2 + 2 × 3x × 5 + (5)2 [Using, (a + b)2 = a2 + 2ab + b2]

= 9x2 + 30x + 25

(ii) (9y – 5) (9y – 5)

= (9y – 5)2

= (9y)2 – 2 × 9y × 5 + (5)2 [Using, (a – b)2 = a2 – 2ab + b2]

= 81y2 – 90y + 25

(iii) (4x + 11y)(4x – 11y)

= (4x)2 – (11y)2

= 16x2 – 121y2 [Using, (a + b)(a – b) = a2 – b2]

(iv) (3m/2 + 2n/3) (3m/2 – 2n/3)

= (3m/2)2 – (2n/3)2

= 9m2/4 – 4n2/9 [Using, (a + b)(a – b) = a2 – b2]

(v) (2/a + 5/b) (2a + 5/b)

= (2/a + 5/b)2

= (2/a)2 + 2(2/a)(5/b) + (5/b)2 [Using, (a + b)2 = a2 + 2ab + b2]

= 4/a2 + 20a/b + 25/b2

(vi) (p2/2 + 2/q2) (p2/2 – 2/q2)

= (p2/2)2 – (2/q2)2 [Using, (a + b)(a – b) = a2 – b2]

= p4/4 – 4/q4

2. Using the identities, evaluate the following:

(i) 812

(ii) 972

(iii) 1052

(iv) 9972

(v) 6.12

(vi) 496 × 504

(vii) 20.5 × 19.5





(viii) 9.62

Solution:

(i) (81)2 = (80 + 1)2

= (80)2 + 2 × 80 × 1 + (1)2 [Using, (a + b)2 = a2 + 2ab + b2]

= 6400 + 160+ 1

= 6561

(ii) (97)2 = (100 – 3)2

= (100)2 – 2 × 100 × 3 + (3)2 [Using, (a – b)2 = a2 – 2ab + b2]

= 10000 – 600 + 9

= 10009 – 600

= 9409

(ii) (105)2 = (100 + 5)2

= (100)2 + 2 × 100 × 5 + (5)2 [Using, (a + b)2 = a2 + 2ab + b2]

= 10000+ 1000 + 25

= 11025

(iv) (997)2 = (1000 – 3)2

= (1000)2 – 2 × 1000 × 3 + (3)2 [Using, (a – b)2 = a2 – 2ab + b2]

= 1000000 – 6000 + 9

= 1000009 – 6000

= 994009

(v) (6.1)2 = (6 + 0.1)2

= (6)2 + 2 × 6 × 0.1 +(0.1)2 [Using, (a + b)2 = a2 + 2ab + b2]

= 36 + 1.2 + 0.01

= 37.21

(vi) 496 × 504

= (500 – 4) (500 + 4) [Using, (a + b) (a – b) = a2 – b2]

= (500)2 – (4)2

= 250000 – 16

= 249984

(vii) 20.5 × 19.5

= (20 + 0.5) (20 – 0.5) [Using, (a + b) (a – b) = a2 – b2]

= (20)2 – (0.5)2

= 400 – 0.25

= 399.75

(viii) (9.6)2 = (10 – 0.4)2

= (10)2 – 2 × 10 × 0.4 + (0.4)2 [Using, (a – b)2 = a2 – 2ab + b2]

= 100 – 8.0 + 0.16

= 92.16





3. Find the following squares, using the identities:

(i) (pq + 5r)2 (ii) (5a/2 – 3b/5)2

(iii) (√2a + √3b)2 (iv) (2x/3y – 3y/2x)2

Solution:

(i) (pq + 5r)2

= (pq)2 + 2 × pq × 5r + (5r)2 [Using, (a + b)2 = a2 + 2ab + b2]

= p2q2 + 10pqr + 25r2

(ii) (5a/2 – 3b/5)2

= (5a/2)2 – 2 × (5a/2) × (-3b/5) + (3b/5)2 [Using, (a – b)2 = a2 – 2ab + b2]

= 25a2/4 – 3ab + 9b2/25

(iii) (√2a + √3b)2

= (√2a)2 + 2 × √2a × √3b + (√3b)2 [Using, (a + b)2 = a2 + 2ab + b2]

= 2a2 + 2√6ab + 3b2

(iv) (2x/3y – 3y/2x)2

4. Using the identity, (x + a) (x + b) = x2 + (a + b)x + ab, find the following products:

(i) (x + 7) (x + 3)

(ii) (3x + 4) (3x – 5)

(iii) (p2 + 2q) (p2 – 3q)

(iv) (abc + 3) (abc – 5)

Solution:

(i) (x + 7) (x + 3)

= (x)2 + (7 + 3)x + 7 × 3

= x2 + 10x + 21

(ii) (3x + 4) (3x – 5)

= (3x)2 + (4 – 5) (3x) + 4 × (-5)

= 9x2 – 3x – 20

(iii) (P2 + 2q)(p2 – 3q)

= (p2)2 + (2q – 3q)p2 + 2q × (-3q)

= p4 – p2q – 6pq

(iv) (abc + 3) (abc – 5)





= (abc)2 + (3 – 5)abc + 3 × (-5)

= a2b2c2 – 2abc – 15

5. Using the identity, (x + a) (x + b) = x2 + (a + b)x + ab, evaluate the following:

(i) 203 × 204

(ii) 8.2 × 8.7

(iii) 107 × 93

Solution:

(i) 203 × 204

= (200 + 3) (200 + 4)

= (200)2 + (3 + 4) × 200 + 3 × 4

= 40000 + 1400 + 12

= 41412

(ii) 8.2 × 8.7

= (8 + 0.2) (8 + 0.7)

= (8)2 + (0.2 + 0.7) × 8 + 0.2 × 0.7

= 64 + 8 × (0.9) + 0.14

= 64 + 7.2 + 0.14

= 71.34

(iii) 107 × 93

= (100 + 7) (100 – 7)

= (100)2 + (7 – 7) × 100 + 7 × (-7)

= 10000 + 0 – 49

= 9951

6. Using the identity a2 – b2 = (a + b) (a – b), find

(i) 532 – 472

(ii) (2.05)2 – (0.95)2

(iii) (14.3)2 – (5.7)2

Solution:

(i) 532 – 472

= (50 + 3) (50 – 3)

= (50)2 – (3)2

= 2500 – 9

= 2491

(ii) (2.05)2 – (0.95)2

= (2.05 + 0.95) (2.05 – 0.95)

= 3 × 1.10

= 3.3

(iii) (14.3)2 – (5.7)2





= (14.3 + 5.7) (14.3 – 5.7)

= 20 × 8.6

= 172

7. Simplify the following:

(i) (2x + 5y)2 + (2x – 5y)2

(ii) (7a/2 – 5b/2)2 – (5a/2 – 7b/2)2

(iii) (p2 – q2r)2 + 2p2q2r

Solution:

(i) (2x + 5y)2 + (2x – 5y)2 [Using, (a ± b)2 = a2 ± 2ab + b2]

= (2x)2 + 2 × 2x × 5y + (5y)2 + (2x)2 – 2 × 2x × 5y + (5y)2

= 4x2 + 20xy + 25y2 + 4x2 – 20xy + 25y2

= 8x2 + 50y2

(ii) (7a/2 – 5b/2)2 – (5a/2 – 7b/2)2

(iii) (p2 – q2r)2 + 2p2q2r [Using, (a – b)2 = a2 – 2ab + b2]

= (p2)2 – 2 × p2 × q2r + (q2r)2 + 2p2q2r

= p4 – 2p2q + q4r2 + 2p2q2r

= p4 + q4r2

8. Show that:

(i) (4x + 7y)2 – (4x – 7y)2 = 112xy

(ii) (3p/7 – 7q/6)2 + pq = 9p2/49 + 49q2/36

(iii) (p – q)(p + q) + (q – r)(q + r) + (r – p) (r + p) = 0

Solution:

(i) Taking LHS, we have

LHS = (4x + 7y)2 – (4x – 7y)2 [Using, (a ± b)2 = a2 ± 2ab + b2]

= [(4x)2 + 2 × 4x × 7y + (7y)2] – [(4x)2 – 2 × 4x + 7y + (7y)2]

= (16x2 + 56xy + 49y2) – (16x2 – 56xy + 49y2)

= l6x2 + 56xy + 49y2 – 16x2 + 56xy – 49y2





= 112xy = RHS

(ii) Taking LHS, we have

(iii) Taking LHS, we have

LHS = (p – q) (p + q) + (q – r) (q + r) + (r – p)(r + p)

= p2 – q2 + q2 – r2 + r2 – p2 [Using, (a + b) (a – b) = a2 – b2]

= 0 = RHS

9. If x + 1/x = 2, evaluate:

(i) x2 + 1/x2 (ii) x4 + 1/x4

Solution:

(i) We have, x + 1/x = 2

On squaring on both sides, we get

(x + 1/x)2 = 22

x2 + 2 × x × 1/x + 1/x2 = 4

x2 + 2 + 1/x2 = 4

x2 + 1/x2 = 4 – 2

Thus,

x2 + 1/x2 = 2

(ii) Again squaring, we get

(x2 + 1/x2)2 = 22

x4 + 2 × x2 × 1/x2 + 1/x4 = 4

x4 + 2 + 1/x4 = 4

x4 + 1/x4 = 4 – 2

Thus,

x4 + 1/x4 = 2

10. If x – 1/x = 7, evaluate:

(i) x2 + 1/x2 (ii) x4 + 1/x4

Solution:

We have, x – 1/x = 7






(x – 1/x)2 = 72

x2 – 2 × x2 × 1/x2 + 1/x2 = 49

x2 – 2 + 1/x2 = 49

x2 + 1/x2 = 49 + 2

Thus,

x2 + 1/x2 = 51

(ii) Again squaring, we get

(x2 + 1/x2)2 = 512

x4 + 1/x4 + 2 × x2 × 1/x2 = 2601

x4 + 1/x4 + 2 = 2601

x4 + 1/x4 = 2601 – 2

Thus,

x4 + 1/x4 = 2599

11. If x2 + 1/x2 = 23, evaluate:

(i) x + 1/x (ii) x – 1/x

Solution:

We have, x2 + 1/x2 = 23

(i) (x + 1/x)2 = x2 + 1/x2 + 2

= 23 + 2

= 25

Taking square root on both sides, we get

(x + 1/x) = ±5

Thus, x + 1/x = 5 or -5

(ii) (x – 1/x)2 = x2 + 1/x2 – 2

= 23 – 2

= 21

Taking square root on both sides, we get

(x + 1/x) = ±√21

Thus, x + 1/x = √21 or -√21

12. If a + b = 9 and ab = 10, find the value of a2 + b2.

Solution:

Given,

a + b = 9 and ab = 10

Now, squaring a + b = 9 on both sides, we have

(a + b)2 = (9)

a2 + b2 + 2ab = 81

a2 + b2 + 2 × 10 = 81

a2 + b2 + 20 = 81

a2 + b2 = 81 – 20 = 61

∴ a2 + b2 = 61





13. If a – b = 6 and a2 + b2 = 42, find the value of

Solution:

Given

a – b = 6 and a2 + b2 = 42

a – b = 6

Now, squaring a – b = 6 on both sides, we have

(a – b)2 = (6)2

a2 + b2 – 2ab = 36

42 – 2ab = 36

2ab = 42 – 36 = 6

ab = 6/2 = 3

∴ ab = 3

14. If a2 + b2 = 41 and ab = 4, find the values of

(i) a + b

(ii) a – b

Solution:

Given, a2 + b2 = 41 and ab = 4

(i) (a + b)2 = a2 + b2 + 2ab

= 41 + 2 × 4

= 41 + 8

= 49

∴ a + b = ±7

(ii) (a – b)2 = a2 + b2 – 2ab

= 41 – 2 × 4

= 41 – 8

= 33

∴ a – b = ±√33





Check Your Progress

1. Add the following expressions:

(i) -5x2y + 3xy2 – 7xy + 8, 12x2y – 5xy2 + 3xy – 2

(ii) 9xy + 3yz – 5zx, 4yz + 9zx – 5y, -5xz + 2x – 5xy

Solution:

(i) (-5x2y + 3xy2 – 7xy + 8) + (12x2y – 5xy2 + 3xy – 2)

= 7x2y – 2xy2 – 4xy + 6

(ii) (9xy + 3yz – 5zx) + (4yz + 9zx – 5y, -5xz + 2x – 5xy)

= 4xy + 7yz – zx + 2x – 5y

2. Subtract:

(i) 5a + 3b + 11c – 2 from 3a + 5b – 9c + 3

(ii) 10x2 – 8y2 + 5y – 3 from 8x2 – 5xy + 2y2 + 5x – 3y

Solution:

(i) 5a – 3b + 11c – 2 from 3a + 5b – 9c + 3

= (3a + 5b – 9c + 3) – (5a – 3b + 11c – 2)

= 3a + 5b – 9c + 3 – 5a + 3b – 11c + 2

= -2a + 8b – 20c + 5

(ii) 10x2 – 8y2 + 5y – 3 from 8x2 – 5xy + 2y2 + 5x – 3y

= (8x2 – 5xy + 2y2 + 5x – 3y) – (10x2 – 8y2 + 5y – 3)

= 8x2 – 5xy + 2y2 + 5x – 3y – 10x2 + 8y2 – 5y + 3

3. What must be added to 5x2 – 3x + 1 to get 3x3 – 7x2 + 8?

Solution:

From the question, the required expression is

= (3x3 – 7x2 + 8) – (5x2 – 3x + 1)

= 3x3 – 7x2 + 8 – 5x2 + 3x – 1

= 3x3 – 12x2 + 3x + 7

4. Find the product of

(i) 3x2y and -4xy2

(ii) –(4/5)xy, (5/7)yz and –(14/9)zx

Solution:

Product of:

(i) 3x2y and -4xy2

= 3x2 × (-4xy2)

= -12x2+1 y1+2

= 12x3y3





(ii) –(4/5)xy, (5/7)yz and –(14/9)zx

= –(4/5)xy × (5/7)yz × –(14/9)zx

= –(4/5) × (5/7) × –(14/9) x2y2z2

= (8/9)x2y2z2

5. Multiply:

(i) (3pq – 4p2 + 5q2 + 7) by -7pq

(ii) (3/4x2y – 4/5xy + 5/6xy2) by – 15xyz

Solution:

(i) (3pq – 4p2 + 5q2 + 7) × (-7pq)

= -7pq × 3pq – 7pq × (-4p2) + (-7pq) (5q2) – 7pq × 7

= -21p2q2 + 28p3q – 35pq3 – 49pq

(ii) (3/4x2y – 4/5xy + 5/6xy2) × (– 15xyz)

6. Multiply:

(i) (5x2 + 4x – 2) by (3 – x – 4x2)

(ii) (7x2 + 12xy – 9y2) by (3x2 – 5xy + 3y2)

Solution:

(i) (5x2 + 4x – 2) × (3 – x – 4x2)

= 5x2(3 – x – 4x2) + 4x(3 – x – 4x2) – 2(3x – x – 4x2)

= 15x2 – 5x3 – 20x4 + 12x – 4x2 – 16x3 – 6x + 2x + 8x2

= -20x4 – 21x3 + 19x2 + 14x – 6

(ii) (7x2 + 12xy – 9y2) x (3x2 – 5xy + 3y2)

= 7x2(3x2 – 5xy + 3y2) + 12xy(3x2 – 5xy + 3y2) – 9y2(3x2 – 5xy + 3y2)

= 21x4 – 35x3y + 21x2y2 + 36x3y – 60x2y2 + 36xy3 – 27x2y2 + 45xy3 – 27y4

= 21x4 + x3y + 81xy3 – 66x2y2 – 27y4

7. Simplify the following expressions and evaluate them as directed:

(i) (3ab – 2a2 + 5b2) x (2b2 – 5ab + 3a2) + 8a3b – 7b4 for a = 1, b = -1

(ii) (1.7x – 2.5y) (2y + 3x + 4) – 7.8x2 – 10y for x = 0, y = 1.

Solution:

(i) (3ab – 2a2 + 5b2) × (2b2 – 5ab + 3a2) + 8a3b – 7b4

= 3ab(2b2 – 5ab + 3a2) – 2a2(2b2 – 5ab + 3a2) + 5b2(2b2 – 5ab + 3a2) + 8a3b – 7b4





= 6ab32 – 15a2b2 + 9a3b – 4a2b2 + 10a3b – 6a4 + 10b4 – 25ab3 + 15a2b2 + 8a3b – 7b4

= 27a3b – 4a2b2 – 19ab3 – 6a4 + 3b4

Putting, a = 1 and b = (-1)

= 27(1 )3 (-1) – 4(1)2 (-1)2 – 19 (1) (-1)3 – 6(1)4 + 3(-1)4

= -27 – 4 + 19 – 6 + 3

= -37 + 22

= -15

(ii) (1.7x – 2.5y) (2y + 3x + 4) – 7.8x2 – 10y

1.7x(2y + 3x + 4) – 2.5y(2y + 3x + 4) – 7.8x2 – 10y

= 3.4xy + 5.1x2 + 6.8x – 5y2 – 7.5xy – 10y – 7.8x2 – 10y

= -2.7x2 – 4.1xy – 5y2 + 6.8x – 20y

Putting, x = 0 and y = 1

= -2.7 × 0 – 4.1 × 0 × 1 – 5(1)2 + 6.8 × 0 – 20 × 1

= 0 + 0 – 5 + 0 – 20

= -25

8. Carry out the following divisions:

(i) 66pq2r3 ÷ 11qr2

(ii) (x3 + 2x2 + 3x) ÷ 2x

Solution:

(i) 66pq2r3/ 11qr2

= 6pq2-1r3-2

= 6pqr

(ii) (x3 + 2x2 + 3x)/ 2x

= x3/2x + 2x2/2x + 3x/2x

= ½ x2 + x + 3/2

9. Divide 10x4 – 19x3 + 17x2 + 15x – 42 by 2x2 – 3x + 5.

Solution:

(10x4 – 19x3 + 17x2 + 15x – 42) ÷ (2x2 – 3x + 5)

Performing long division, we have





Thus, Quotient = 5x2 – 2x – 7 and Remainder = 4x – 7

10. Using identities, find the following products:

(i) (3x + 4y) (3x + 4y)

(ii) (5a/2 - b) (5a/2 – b)

(iii) (3.5m – 1.5n) (3.5m + 1.5n)

(iv) (7xy – 2) (7xy + 7)

Solution:

(i) (3x + 4y) (3x + 4y)

= (3x + 4y)2

= (3x)2 + 2 × 3x × 4y + (4y)2 [Using, (a + b)2 = a2 + 2ab + b2]

= 9x2 + 24xy + 16y2

(ii) (5a/2 - b) (5a/2 – b)

= (5a/2 - b)2

= (5a/2)2 + 2 × 5a/2 × (-b) + (b)2 [Using, (a – b)2 = a2 – 2ab + b2]

= 25a2/4 – 5ab + b2

(iii) (3.5m – 1.5n) (3.5m + 1.5n)

= (3.5m)2 – (1.5n)2 [Using, (a – b)(a + b) = a2 – b2]

= 12.25m2 – 2.25n2

(iv) (7xy – 2)(7xy + 7)

= (7xy)2 + (-2 + 7) × (7xy) + (-2) × 7 [Using, (x + a)(x + b) = x2 + (a + b)x + ab]

= 49x2y2 + 35xy – 14

11. Using suitable identities, evaluate the following:

(i) 1052

(ii) 972

(iii) 201 × 199

(iv) 872 – 132

(v) 105 × 107

Solution:

(i) (105)2 = (100 + 5)2

= (100)2 + 2 × 100 × 5 + (5)2 [Using, (a + b)2 = a2 + 2ab + b2]

= 10000 + 1000 + 25

= 11025

(ii) (97)2 = (100 – 3)2

= (100)2 – 2 × 100 × 3 + (3)2 [Using, (a – b)2 = a2 – 2ab + b2]

= 10000 – 600 + 9

= 10009 – 600

= 9409





(iii) 201 × 199 = (200 + 1) (200 – 1)

= (200)2 – (1)2 [Using, (a + b) (a – b) = a2 – b2]

= 40000 – 1

= 39999

(iv) 872 – 132

= (87 + 13) (87- 13) [Using, a2 – b2 = (a + b)(a – b)]

= 100 × 74

= 7400

(v) 105 × 107

= (100 + 5) (100 + 7)

= (100)2 + (5 + 7) × 100 + 5 × 7 [Using, (x + a)(x – b) = x2 + (a + b)x + ab]

= 10000 + 1200 + 35

= 11235

12. Prove that following:

(i) (a + b)2 – (a – b)2 + 4ab

(ii) (2a + 3b)2 + (2a – 3b)2 = 8a2 + 18b2

Solution:

(i) Taking the RHS, we have

RHS = (a – b)2 + 4ab

= a2– 2ab + b2 + 4ab

= a2 + 2ab + b2

= (a + b)2 = L.H.S.

(ii) Taking the LHS, we have

LHS = (2a + 3b)2 + (1a – 3b)2

= (2a)2 + 2 × 2a × 3b + (3b)2 + (2a)2 – 2 × 2a × 3b + (3b)2

= 4a2 + 12ab + 9b2 + 4a2 – 12ab + 9b2

= 8a2 + 18b2 = RHS

13. If x + 1/x = 5, evaluate

(i) x2 + 1/x2 (ii) x4 + 1/x4

Solution:

(i) We have, x + 1/x = 5


(x + 1/x)2 = 52

x2 + 1/x2 + 2 × x × 1/x = 25

x2 + 2 + 1/x2 = 25

x2 + 1/x2 = 25 – 2

Hence, x2 + 1/x2 = 23

(ii) Again, squaring x2 + 1/x2 = 23 on both sides, we get





(x2 + 1/x2)2 = 232

x4 + 1/x4 + 2 × x4 × 1/x4 = 529

x4 + 1/x4 + 2 = 529

x4+ 1/x4 = 529 – 2

Hence,

x4 + 1/x4 = 527

14. If a + b = 5 and a2 + b2 = 13, find ab.

Solution:

Given,

a + b = 5 and a2 + b2 = 13

On squaring a + b = 5 both sides, we get

(a + b)2 = (5)2

a2 + b2 + 2ab = 25

13 + 2ab = 25 ⇒ 2ab = 25 – 13 = 12

⇒ ab = 12/2 = 6

∴ ab = 6



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